# Difference between quantiles of different random variables

Let $Q_{\alpha}(X)$ for $\alpha \in (0,1)$ be the $\alpha$-th quantile of the random variable $X$. Suppose that $X$ and another random variable $Y$ (independent of $X$) have nice continuous (or maybe even smooth) densities. Can I get an upper bound on the difference: $$\left| Q_{\alpha}(X+Y)- Q_{\alpha}(X) \right|$$ in terms of the densities of $X$ and $Y$?

## 1 Answer

Basic Approach. Note that we can write

$$Q_\alpha(X) = F_X^{-1}(\alpha)$$

where $$F_X(x) \equiv P(X < x)$$ is the cumulative distribution function (CDF) of $$X$$. So the expression in the absolute value is

$$F_{X+Y}^{-1}(\alpha)-F_X^{-1}(\alpha)$$

where the corresponding PDF

$$f_{X+Y} = f_X \ast f_Y$$

That is, $$f_{X+Y}$$ is the convolution of $$f_X$$ and $$f_Y$$. (Thanks to A.S. in the comments for catching this, and shame on me for not noticing for over four years!)

Does that help? Or do you need something more specific?

• But $F_{X+Y}\neq F_X*F_Y$. pdfs, not cdfs, convolute. – A.S. Oct 10 '15 at 5:41
• @A.S.: I'm astounded that I didn't see this comment (or the mistake that precipitated it). Thanks for the catch, and I'll edit. – Brian Tung Jan 25 '20 at 0:09